Find the unit vector in the direction of $\vec{v}=\left( 12, -5 \right)$. $($ $~,$ $)$
Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( 12, -5 \right) }{\sqrt{12^2+(-5)^2}} \\\\\\ &= \dfrac{1}{\sqrt{169}} \cdot \left( 12, -5 \right) \\\\\\ &= \left( {\dfrac{12}{13}}, {-\dfrac{5}{13}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {\dfrac{12}{13}} \right)^2 + \left( {-\dfrac{5}{13}} \right)^2} \\\\\\ &= \sqrt{\dfrac{144}{169} + \dfrac{25}{169}} \\\\\\ &= \sqrt{\dfrac{169}{169}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {\dfrac{12}{13}}, {-\dfrac{5}{13}}\right) $ Visualizing the answer: $12$ $-5$ $\dfrac{12}{13}$ $\dfrac{-5}{13}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$.